Question

If a wire is stretched, so that its length $20\%$ more than its initial length, the percentage increase in the resistance of the wire is-

• A. $40\%$
• B. $10\%$
• C. $44\%$
• D. $25\%$

Answer 1

The correct option is C $44\%$

Given, the wire is stretched, but its volume remains the same.

$V_i=V_f$

$A_1{l}_1=A_2{l}_2\Rightarrow \frac{l_1}{l_2}=\frac{A_2}{A_1}$

Resistance of a wire is, $R=\left(\frac{\rho l}{A}\right)$

$R\propto \frac{l}{A}$

$\frac{R_1}{R_2}=\left(\frac{l_1}{l_2}\right)\times \left(\frac{A_2}{A_1}\right)$

$\frac{R_1}{R_2}={\left(\frac{l_1}{l_2}\right)}^2$

Here, $l_1=l$ and $l_2=1.2l$

$R_2=\frac{R_1}{{\left(\frac{l_1}{l_2}\right)}^2}=1.44R_1$

Percentage increase in the reistance is,

$\%\text{increase}=\frac{R_2-R_1}{R_1}\times 100$

$=\frac{(1.44-1)R_1}{R_1}\times 100=44\%$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(C\right)$ is the correct answer.